Subject: Chemistry — Ionic Equilibrium
Question: A solution is prepared by mixing 0.01 mole each of
$H_{2}CO_{3},NaHCO_{3},Na_{2}CO_{3}$ and $NaOH$ in 100 mL of water. pH
of the resulting solution is \_\_\_\_\_\_\_\_\_\_.
\[Given: $pKa_{1}$ and $pKa_{2}$ of $H_{2}CO_{3}$ are 6.37 and 10.32,
respectively; $\log 2 = 0.30$\]
Answer: 10.02
Solution: $n_{H_{2}CO_{3}} = n_{NaHCO_{3}} = n_{Na_{2}CO_{3}} = n_{NaOH} = 0.01\ mole$.
$H_{2}CO_{3}$ $+$ $NaOH$ $\rightarrow$ $NaHCO_{3}$
0.01 mole 0.01 mole 0.01 mole
0 0 0.02 mole
$NaHCO_{3}$ $+$ $Na_{2}co_{3}$
0.02 mole 0.01 mole
this act as acidic buffer solution.
$$pH = pKa_{2} + \log\frac{\left\lbrack Na_{2}CO_{3} \right\rbrack}{\left\lbrack NaHCO_{3} \right\rbrack}$$
$$= 10.32 + \log\left( \frac{0.01}{0.02} \right)$$
$$= 10.32 \times 0.3$$
$$pH = 10.02$$